I recently had to prove that $$\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A})\det(\mathbf{B})$$

I thought that it would be an interesting exercise to show this using the standard definition of the determinant of a matrix that physicists usually give (using Einstein summation):

$$\det(\mathbf{A}) \triangleq \varepsilon^{i_{1}\cdots i_{n}} a_{1 i_{1}} \cdots a_{n i_{n}}$$

We note that if we denote $\mathbf{C} = \mathbf{A}\mathbf{B}$, then we have $c_{ij} = a_{ik}b_{kj}$, and thus:

$$\begin{align*}\det(\mathbf{C}) &= \varepsilon^{i_{1}\cdots i_{n}}c_{1 i_{1}}\cdots c_{n i_{n}} \\

&= \varepsilon^{i_{1}\cdots i_{n}} a_{1 k_{1}}b_{k_{1} i_{1}} \cdots a_{n k_{n}}b_{k_{n} i_{n}} \\

&= a_{1 k_{1}}\cdots a_{n k_{n}} \varepsilon^{i_{1} \cdots i_{n}}b_{k_{1} i_{1}}\cdots b_{k_{n} i_{n}}\end{align*}$$

We note that the second half of this formula is the determinant of some matrix $\mathbf{B’}$, which looks something like:

$$\det(\mathbf{B’}) = \begin{vmatrix}\longleftarrow & \vec{b_{k_{1}}} & \longrightarrow \\

& \vdots & \\

\longleftarrow & \vec{b_{k_{n}}} & \longrightarrow\end{vmatrix},$$

Where $b_{k_{i}}$ is the $k_{i}$th row vector of $\mathbf{B}$. We can thus see that the only valid $\vec{k} = \{k_{1},\dots,k_{n}\}$ are given by permutations of $\{1,\dots,n\}$. We note that we have the general identity: If $\mathbf{A}_{P}$ is a matrix with it’s row’s permuted by some permutation $P = \{P(1),\dots,P(n)\}$, we have that the determinant is multiplied by the signature of $P$, that is:

$$\mathbf{A}_{P}=\begin{pmatrix}\longleftarrow & \vec{a_{P(1)}} & \longrightarrow \\ & \vdots & \\ \longleftarrow & \vec{a_{P(n)}} & \longrightarrow\end{pmatrix} \implies \det(\mathbf{A}_{P}) = \operatorname{sgn}(P)\det(\mathbf{A})$$

Thus we have:

$$\begin{align*}\det(\mathbf{A}\mathbf{B}) &= \varepsilon^{k_{1}\cdots k_{n}}a_{1 k_{1}}\cdots a_{n k_{n}} \varepsilon^{i_{1}\cdots i_{n}}b_{k_{1} i_{1}} \cdots b_{k_{n} i_{n}} \\

&= \det(\mathbf{A})\det(\mathbf{B})\end{align*}$$

Q.E.D.